A counterflow heat exchanger is being used to cool a flow of hot water using a c
ID: 2321267 • Letter: A
Question
A counterflow heat exchanger is being used to cool a flow of hot water using a cold 20% ethylene glycol solution. The hot water enters at 80 degree F with a volumetric flow rate of 400 gpm. The ethylene glycol solution enters the heat exchanger at -10 degree F and leaves at 15 degree F. The ethylene glycol solution is flowing at a volumetric flow rate of 300 gpm. The overall heat transfer coefficient for the heat exchanger is found to be 155 Btu/h.ft^2.degree F. Determine the following: Outlet temperature of the water (degree F) Heat transfer rate between the water and ethylene glycol solution (Btu/h) Required heat transfer area of the heat exchanger (ft^2) Solve Problem 5.12 for a parallel-flow heat exchanger. Solve Problem 5.12 for a STHX with 1-shell pass and 2-tube passes. Problem 5.12 for a CFHX where both fluids are unmixed.Explanation / Answer
1)OUT LET TEMPERATURE OF WATER
ON CONVERSION GPM TO kg/Hr
MASS FLOW RATE OF HOT WATER=400GPM=90909Kg/Hr
1kg/hr=0.0044gpm
MASS FLOW RATE OF COLD ETHYLENE GLYCOL=300GPM=71748Kg/Hr
HOT WATER INLET TEMPERATURE=80 DEGREE FAHREIGHNHEAT=26.6DEGREE CENTIGRADE
ETHYLENE GLYCOL INLET TEMPERATURE=-10DEGREE FAHREIGHNHEAT=-23.5 DEGREECENTIGRADE
ETHYLENE GLYCOL OUTLET TEMPERATURE=15DEGREE FAHREIGHNHEAT=-9.5 DEGREECENTIGRADE
A)HOT WATER OUT LET TEMPERATURE=?
FROM ENERGY BALANCE EQUATION
MH*CH*(THI-THO)=MC*CC*(TCO-TCI)
MC=MASS OF COLD FLUID=ETHYLENE GLYCOL
CC=SPECIFIC HEAT OF COLD FLUID=202KJ/KG
90909*4.2*(26.06- THO) =1748*2.2*(-9.5-(-23.5))
THO = HOT WATER OUT LET TEMPERATURE=20.9 DEGREE CENTIGRADE
B)HEAT EXCHANGE BETWEEN WATER AND ETHYLENE GLYCOL
Q= MH*CH*(THI-THO) =90909*4.2*(26.06- 20.9 =604544.15watts
C)AREA OF HEAT EXCHANGER FOR SHELL AND TUBE WITH 1 SHELL PASS AND 2 TUBE PASS
Q=U*A*LMTD
LMTD=2- 1/ln(2/ 1)
1=THo-TcO
2=THi-TcO
LMTD=26.06-(-9.5)-(20.9-(-23.5))/ln(26.6-(-9.5))/20.9-(-23.5))
=401 degree centigrade =674 degree fahreighn heat
AREA OF HEAT EXCHANGER=Q/U*LMTD
U=OVER ALL HEAT TRANSFER=155BTU/h.FT2 0F =1.92 W/M2 0C
AREA=604544.15/674*1.92=467.17M2
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